How to get a stand from JoJo

The average number of hurricanes per globally stands at 15 hurricanes per year due to excess heat built up in Earth’s atmosphere.

Hurricanes are huge storms produced in the tropics due to large amounts of water vapor present in the atmosphere from the ocean coupled with strong winds with devastating effects on land.

The activities of man such as burning of fossil fuels have resulted in increased build of heat in the atmosphere and warming of the oceans. This is known as global warming.

Due to global warming, on average, the number of hurricanes per globally stands at 15 hurricanes per year.

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Answer:

a) a = 2.35 m/s^2

Explanation:

(a) In order to calculate the magnitude of the acceleration of the ball, you use the following formula, for the position of the ball:

\(x=v_ot+\frac{1}{2}at^2\)     (1)

x: position of the ball after t seconds = 87 m

t: time  = 8.6 s

a: acceleration of the ball = ?

vo: initial velocity of the ball = 0 m/s

You solve the equation (1) for a:

\(x=0+\frac{1}{2}at^2\\\\a=\frac{2x}{t^2}\)

You replace the values of the parameters in the previous equation:

\(a=\frac{2(87m)}{(8.6s)^2}=2.35\frac{m}{s^2}\)

The acceleration of the ball is 2.35 m/s^2

Answer:

Et = 30018 [W]

Explanation:

To solve this problem we must use the definition of potential energy, which is expressed by the following equation:

\(E_{p} =m*g*h\)

where:

Ep = potencial energy [W]

g = gravity acceleration [m/s^2]

h = elevation = 120 [m]

m = mass flow = [1800/min]

But in order to get watts into potential energy, we must convert the mass flow from kilograms per minute to kilograms per second.

\(1800[\frac{kg}{min} ]*\frac{1}{60}\frac{min}{s}=30[\frac{kg}{s} ]\)

Ep = 30*9.81*120

Ep = 35316 [W]

If we use  85% of the potential  energy will have:

Et = 35316*0.85

Et = 30018 [W]

The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.

The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.

The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.

The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.

The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).

The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).

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This question is not complete, the complete question is;

An electron remains suspended between the surface of the Earth (assumed neutral) and a fixed positive point charge, at a distance of 5.3 m from the point charge. Determine the charge required for this to happen.  The acceleration due to gravity is 9.8 m/s² and the Coulomb constant is 8.98755 x 10⁹ N.m²/C².  Answer in units of C.

Answer: the charge required for this to happen is 1.7415 × 10¹⁹ C

Explanation:

Given the data in the question;

we know that gravitational force on electro = M×g

mass of electron m = 9.10938356 × 10⁻³¹ kilograms

charge of electron e = 1.60217662 × 10⁻¹⁹ coulombs

given that distance r = 5.3m and the Coulomb constant = 8.98755 x 10⁹ N.m²/C²

g = 9.8 m/s²

Now for the electron to be suspended, This will happen only if the electrostatic force is equal to weight of electron;

[ (1/4πε0) × q × e] / r²   =   m × g

so we substitute

8.98755 x 10⁹ × q × 1.60217662 × 10⁻¹⁹/[5.3]² = 9.10938356 × 10⁻³¹ × 9.8

q =  8.9271 × 10⁻³⁰ / 5.1262 × 10⁻¹¹

q = 1.7415 × 10¹⁹ C

Therefore the charge required for this to happen is 1.7415 × 10¹⁹ C

[1.43996 × 10⁹ × q] / 29.09 =  8.92719 x 10⁻³⁰

1.43996 × 10⁹ × q  = 2.5969 × 10⁻²⁶

q = 2.5969 × 10⁻²⁶ / 1.43996 × 10⁹

q =  

One of the conclusion about the history of sports that can be drawn from the movie "Champions" is that sports have a long and rich history that is intertwined with the history of human civilization.

From the ancient Olympic Games to the modern-day Olympics, sports have been a way for people to come together and compete in a spirit of sportsmanship and competition. Sports have also been a way for people to express themselves, to build community, and to achieve personal goals.

The movie "Champions" tells the story of a group of young men who come together to form a football team. The team is made up of boys from different backgrounds and with different abilities. However, they are all united by their love of football and their desire to win.

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Answer:

\(f(x=3.00m)=0.075mcos(\frac{2\pi(2.00Hz)}{v}(3.00m))\)

Explanation:

To find the equation of the wave you use the general equation for a wave, given by:

\(f(x)=Acos(k x-\omega t)\)

A: amplitude of the wave = 0.075m

k: wave number

you select a cosine function because for x=0 and t= 0 you get a maximum displacement.

To find the displacement of the wave for x=0 you can consider that the form of the wave is independent of time t.

Then, you calculate k:

\(k=\frac{\omega}{v}=\frac{2\pi f}{v}\)

Thus, you need the value of the speed of the wave (you only have the frequency f), in order to calculate f(x), for x=3.00m:

\(f(x=3.00m)=0.075mcos(\frac{2\pi(2.00Hz)}{v}(3.00m))\)

Over the time period from t₁ to just before t₃, the two-block-spring system is linear momentum does not change because there are no external forces acting on the system to change its momentum.

Explain why the linear momentum of the two-block–spring system is either constant or not during the time period between t₁ and the time just before t₃?

The linear momentum of the two-block-spring system remains constant from t₁ until just before t₃.

Prior to the collision at time t₂, block A has a momentum to the right that is Pa = ma ×va, whereas block B has a momentum that is Pb = 0 because it is at rest. Because there is no external force acting on the system, its overall momentum is conserved.

All out Ptotal = Dad + Pb = mama × va

Expect there is immaterial erosion between each block and the surface. Take into consideration the spring, block A, and the system for parts (a) to (d).

Following the collision at time t₂, block B experiences a velocity of vb to the right and comes to a stop. Because the collision is elastic, all of the system's kinetic energy is conserved, but momentum is still conserved:

The two-block system is moving to the right at a constant velocity vb from time t₂ to just before time t₃, so the momentum of the system is constant. Ptotal = ma₀ plus mbvb plus mb ×vb

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Since Taylor has set this fitness goal: I will go to the YMCA and lift weights for 30 minutes every other day for the entire month of January. The option that MOST likely TRUE about this goal is option A:It is a short-term goal.

A short-term goal is a goal that can be achieved within a relatively short period of time, usually less than a year. In this case, the goal is to go to the YMCA and lift weights for 30 minutes every other day for the entire month of January.

In this case, Taylor's goal is to go to the YMCA and lift weights for 30 minutes every other day for the entire month of January. This goal is specific, as it clearly states what Taylor wants to accomplish, and it is measurable.

This goal can be achieved within the month of January and it's a specific, measurable and time-bound goal, which are characteristics of short-term goals.

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See full question below

Taylor has set this fitness goal: I will go to the YMCA and lift weights for 30 minutes every other day for the entire month of January. What is MOST likely TRUE about this goal?

It is a short-term goal.

It is a long-term goal.

It is an unrealistic goal.

It is a mental health goal.

Answer:it’s a short-term goal

Explanation:

because it’s only for the month of January she will be doing this goal.

Answer:

the answer is a....,.......

Answer:

Explanation:

Expression for time period of pendulum is given as follows

\(T=2\pi\sqrt{\frac{l}{g} }\)

where l is length of pendulum and g is acceleration due to gravity .

Putting the given values for first place

\(2.67=2\pi\sqrt{\frac{l}{9.77} }\)

Putting the values for second place

\(T=2\pi\sqrt{\frac{l}{9.81} }\)

Dividing these two equation

\(\frac{T}{2.67} =\sqrt{\frac{9.77}{9.81} }\)

T = 2.66455 s.

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

(a) 0.277 m

(b) 0.553 m

Explanation:

(a) For the closed cylinder,

Applying,

f' = v/4l............. Equation 1

Where f' = fundamental frequency, v = velocity of sound, l = lenght of the cylinder

make l the subject of the equation

l = v/4f'............. Equation 2

From the question,

Given: v = 343 m/s, f' = 310 Hz

Substitute these values into equation 2

l = 343/(4×310)

l = 0.277 m

(b) For the open cylinder,

f' = v/2l

l = v/2f'................ Equation 3

Substitute into equation 3

l = 343/(2×310)

l = 0.553 m

Answer:

a)  \(l=0.276m\)

b)  \(l'=1.11m\)

Explanation:

From the question we are told that:

Frequency \(F=310Hz\)

Speed \(V=343m/s\)

a)

Generally the equation for Frequency is mathematically given by

The cylinder is closed at one end

\(F=\frac{V}{4*l}\)

\(310=\frac{343}{4*l}\)

\(l=0.276m\)

b)

Generally the equation for Frequency is mathematically given by

the cylinder is open at both ends.

\(F=\frac{V}{l'}\)

\(310=\frac{343}{l'}\)

\(l'=1.11m\)

Answer:

Its  the temperature and density of that element in the star

Explanation:

In "The Argonauts," Robinson effectively utilizes language techniques such as metaphor, repetition, and sentence structure to develop his argument to his younger self and engage young readers in the present. Through these techniques, Robinson creates a powerful and relatable narrative that resonates with his audience.

Robinson employs metaphors to convey complex ideas in a compelling and accessible manner. For instance, he compares his struggle with identity and gender to the mythical journey of the Argonauts, making it relatable and captivating for young readers. This metaphorical language enables readers to grasp the profound emotions and challenges he faced during his own personal journey.

Repetition is another technique Robinson employs to reinforce key ideas and create a rhythmic and memorable reading experience. By repeating certain phrases or concepts, he emphasizes their significance and invites readers to reflect on them. This repetition serves to engage young readers by encouraging them to contemplate their own experiences and perspectives.

Furthermore, Robinson carefully structures his sentences to create a sense of rhythm and flow, enhancing the overall readability and impact of his argument. Short, concise sentences create moments of clarity and emphasis, while longer, more descriptive sentences evoke a contemplative and introspective tone. This varied sentence structure adds depth and nuance to his narrative, captivating young readers and keeping them engaged throughout.

In conclusion, through the effective use of metaphor, repetition, and sentence structure, Robinson engages and captivates young readers, inviting them to reflect on their own identities and experiences. His language choices not only develop his argument to his younger self but also establish a connection with present-day young readers, making his work both impactful and relatable.

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The magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹ (Option E).

The electric flux through a surface is given by the product of the electric field strength and the area of the surface projected perpendicular to the electric field.

In this case, the electric field strength is 18 N C⁻¹, and the area of the sheet projected perpendicular to the electric field is 0.450 m²

(since the normal to the sheet makes an angle of 60° with the electric field). Multiplying these values gives the electric flux:

Electric flux = Electric field strength × Area

Electric flux = 18 N C⁻¹ × 0.450 m²

Electric flux = 8.1 N m² C⁻¹

In summary, the magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹. This value is obtained by multiplying the given electric field strength by the projected area of the sheet perpendicular to the electric field.

The angle of 60° is taken into account to determine the effective area for calculating the flux.(Option E).

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Answer:

I and II

Explanation:

They are made of metal so they are great conductors

Answer:

l and ll

Explanation:

Answer:

Alkali Metals

Explanation:

Group 1A — The Alkali Metals. Group 1A (or IA) of the periodic table are the alkali metals: hydrogen (H), lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). These are (except for hydrogen) soft, shiny, low-melting, highly reactive metals, which tarnish when exposed to air.

Answer:

Group 1A (or IA) of the periodic table are the alkali metals:  hydrogen (H), lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr).  These are (except for hydrogen) soft, shiny, low-melting, highly reactive metals, which tarnish when exposed to air.  The name comes from the fact that when these metals or their oxides are dissolved in water, a basic (alkaline) solution results.  Because the alkali metals are very reactive, they are seldom (if ever) found in their elemental form in nature, and are usually found as ionic compounds (except for hydrogen).

The number of food particles in an hour will be 7200.

Expression in maths is defined as the collection of numbers variables and functions by using signs like addition, subtraction, multiplication and division.

Given that the sediment-water interface pumps approximately 5ml of water each minute and filters out of it its nutritious particles. Assuming a concentration of food particles of 40 per cm³,

The number of particles will be calculated below,

Number = ( 40 x 5 ) Per minute

Number o particles = 120 x 60 particles.

Number of particles = 7200 particles

Therefore, the number of food particles in an hour will be 7200.

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A)The total mechanical energy of the two blocks prior to being released from rest can be found by adding the gravitational potential energy of mA and the pulley to zero.

B).The gravitational potential energy of mB and the pulley is(3.30 kg + mp) × 9.81 m/s² × 0 m = 0 J,where mp is the mass of the pulley.The total mechanical energy of the two blocks prior to being released from rest is54.33 J + 0 J = 54.33 J

C) The rotational kinetic energy of the pulley just before mA hits the ground is(0.178 mp) J.

D) The mass of the pulley ismp = (1/2)mpr²/R² =(1/2)(0.020 kg)(0.100 m)²/(0.200 m)² = 0.001 kg = 1 g.r = (1/2)R.

The Atwood's machine shown in Figure 1 consists of two masses mA = 6.50 kg and mB = 3.30 kg. The height of mA above the floor is 0.865 m. When mA hits the floor, its velocity is 1.89 m/s. The pulley has a moment of inertia (1/2)mpr². We have to find the total mechanical energy of the two blocks before they are released, the total mechanical energy when mA hits the ground, the rotational kinetic energy of the pulley just before mA hits the ground, and the mass of the pulley. Let's solve these one by one. Part A The total mechanical energy of the two blocks prior to being released from rest can be found by adding the gravitational potential energy of mA and the pulley to zero.

The equation for gravitational potential energy is mgh. The gravitational potential energy of mA and mB is mAg(h-hB)where h is the height of mA above the floor and hB is the height of mB above the floor. Since the pulley is at the same height as mB, its gravitational potential energy ismBg(h-hB).The gravitational potential energy of mA is6.50 kg × 9.81 m/s² × 0.865 m = 54.33 J.The gravitational potential energy of mB and the pulley is(3.30 kg + mp) × 9.81 m/s² × 0 m = 0 J,where mp is the mass of the pulley.The total mechanical energy of the two blocks prior to being released from rest is54.33 J + 0 J = 54.33 J.Part BThe total mechanical energy of the two blocks when mA hits the ground can be found by adding the kinetic energy of mA, the kinetic energy of mB, and the rotational kinetic energy of the pulley to the gravitational potential energy of mB and the pulley. The equation for kinetic energy is (1/2)mv². The kinetic energy of mA is(1/2) × 6.50 kg × (1.89 m/s)² = 11.54 J.The kinetic energy of mB is(1/2) × 3.30 kg × 0 m/s² = 0 J, since it is at rest.The gravitational potential energy of mB and the pulley is(3.30 kg + mp) × 9.81 m/s² × 0 m = 0 J.The rotational kinetic energy of the pulley is(1/2) × (1/2)mp × R² × ω²,where R is the radius of the pulley and ω is its angular velocity just before mA hits the ground. We can use the fact that the linear speed of the rope is the same on both sides of the pulley to find ω. The equation for linear speed is v = Rω. When mA hits the ground, its speed is 1.89 m/s. The speed of mB is zero. Since the rope is inextensible, the speed of the rope is also 1.89 m/s.

Therefore, the speed of the pulley is also 1.89 m/s. We can find the angular velocity of the pulley by dividing the linear velocity by the radius.ω = v/R = 1.89 m/s ÷ (0.200 m/2) = 18.9 rad/s.The rotational kinetic energy of the pulley is(1/2) × (1/2)mp × R² × ω² =(1/4)mpR²ω² =(1/4)mp(0.200 m)²(18.9 rad/s)² =(0.178 mp) J.The total mechanical energy of the two blocks when mA hits the ground is11.54 J + 0 J + 0 J + (0.178 mp) J = 11.72 J + (0.178 mp) J.Part CThe rotational kinetic energy of the pulley just before mA hits the ground is(0.178 mp) J.Part DWe can find the mass of the pulley by using the moment of inertia of a disk and the mass of the pulley. The moment of inertia of a disk is (1/2)mr². Therefore,(1/2)mpR² = (1/2)mpr²,where R is the radius of the pulley and r is the radius of gyration of the pulley. The radius of gyration of a disk is (1/2)R.

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Answer:

Explanation:

Assuming there is no friction or other force involved, recall that energy is conserved in a system as long as no external force acts on the system.

Using the data from point C, we can find out that the total energy of the system is 156 because  \(E = K+Pe\).

Since at point A the object doesn't move, it has Kinetic energy of 0, because \(K=\frac{1}{2} mv^2\), therefore \(0=\frac{1}{2} m0^2\). However at point A it has maximum Potential energy, because \(Pe=mgh\).

At point B, we can find the Kinetic energy by using \(E = K+Pe\). Substitute values:

\(156=104+K\\52=K\)

At point D, the object has maximum kinetic energy and no potential energy, therefore it's the opposite of point A.

Answer:

Kinetic energy is energy an object possesses due to its motion.

Kinetic energy equation: \(KE = \frac{1}{2} mv^{2}\)

The faster and heavier an object is, the higher the kinetic energy.

Potential magnetic energy is the energy of an object that is able to do work because of its position in a magnetic field.

Answer:

Explanation:

The force of attraction between 2 masses.

The constant acceleration of the train is 50/9 m/s².

The two balls will collide at a height of approximately 10.204 meters above the ground.

Using the kinematic equations of motion, we have:

distance = initial velocity * time + 1/2 * acceleration * time^2

For the first phase of acceleration, the initial velocity is zero, the time is 5 minutes = 300 seconds, and the distance traveled is unknown. So we have:

d1 = 0 + 1/2 * a * (300)^2

For the second phase of constant speed, the initial velocity is v, the time is 5 minutes = 300 seconds, and the distance traveled is also unknown. So we have:

d2 = v * 300

For the third phase of deceleration, the initial velocity is v, the time is also 5 minutes = 300 seconds, and the distance traveled is again unknown. So we have:

d3 = v * 300 + 1/2 * (-a) * (300)^2

The total distance traveled is the sum of these three distances:

distance = d1 + d2 + d3 = 1/2 * a * (300)^2 + v * 600 - 1/2 * a * (300)^2 = v * 600

Since the total distance traveled is given as 10 km = 10000 m, we have:

v * 600 = 10000

Solving for v, we get:

v = 10000/600 = 50/3 m/s

Now we can use the second equation above to find a:

d2 = v * 300 = (50/3) * 300 = 5000 m

Therefore, the constant acceleration of the train is:

a = 2 * (5000 - 1/2 * a * (300)^2) / (300)^2 = 50/9 m/s^2

The constant acceleration of the train is 50/9 m/s^2.

Problem 2: The height of the first ball dropped is given as 10m. Let's assume the height of the collision point is h meters above the ground.

Using the kinematic equation for free fall, we have:

h = 10 + 1/2 * g * t^2

where g is the acceleration due to gravity, which is approximately 9.81 m/s^2, and t is the time it takes for the second ball to reach the collision point after being thrown upwards.

The initial upward velocity of the second ball is 10 m/s, and we know that at the collision point, its velocity will be zero, since it will have reached its maximum height and will be momentarily at rest before falling back down.

Using the kinematic equation for motion with constant acceleration, we have:

0 = 10 + (-g) * t

Solving for t, we get:

t = 10/g = 10/9.81 seconds

Substituting this value of t into the first equation, we get:

h = 10 + 1/2 * 9.81 * (10/9.81)^2

Simplifying, we get:

h = 10.204 m

The two balls will collide at a height of approximately 10.204 meters above the ground.

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Calculating the mass of the block requires a bit of work. The formula for the volume of a rectangular solid is V = l*w*h, where V is the volume, l is the length, w is the width, and h is the height. Using the dimensions given, we can calculate the volume of the block as 30*80*60 = 144000 cubic centimeters.

The density of lead is approximately 11.34 grams per cubic centimeter. To calculate the mass of the block, we can use the formula m = V*d, where m is the mass, V is the volume, and d is the density. Plugging in the values we get m = 144000*11.34 = 1,634,400 grams or approximately 1.63 metric tons.

So, the mass of the block of lead is approximately 1.63 metric tons.

The terminal velocity of the block is approximately 2.08 m/s when it is traveling downward with a drag coefficient of 0.822.

How to calculate terminal velocity?

To calculate the terminal velocity of the block, we need to use the equation for terminal velocity which is:

v_terminal = sqrt((2mg)/(ρAC_d))

where:

First, let's convert the dimensions of the block from centimeters to meters:

width = 16.0 cm = 0.16 m

length = 19.0 cm = 0.19 m

height = 25.0 cm = 0.25 m

Next, let's calculate the cross-sectional area of the block:

A = width * length = 0.16 m * 0.19 m = 0.0304 m^2

Now we can substitute the values into the equation for terminal velocity:

v_terminal = sqrt((24.55 kg9.81 m/s^2)/(1.43 kg/m^30.0304 m^20.822))

v_terminal = 2.08 m/s

Therefore, the terminal velocity of the block is approximately 2.08 m/s when it is traveling downward with a drag coefficient of 0.822.

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The primary difference is that electromagnetic waves can propagate through a vacuum or empty space, while mechanical waves require a physical medium to transmit energy.

A primary difference between an electromagnetic wave and a mechanical wave is the medium through which they propagate.

Electromagnetic waves can propagate through a vacuum or empty space without requiring a material medium. They are generated by the oscillation and interaction of electric and magnetic fields.

Examples of electromagnetic waves include radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. These waves can travel through space, air, or other materials, as they do not rely on physical particles to transmit energy.

On the other hand, mechanical waves require a physical medium to propagate. They are disturbances that travel through a material medium, transferring energy from one location to another. Mechanical waves rely on the interaction and displacement of particles within the medium to transmit energy.

Examples of mechanical waves include sound waves, water waves, seismic waves, and waves on a string. These waves cannot travel through a vacuum as they depend on the physical presence and interaction of particles within the medium.

In summary, the primary difference is that electromagnetic waves can propagate through a vacuum or empty space, while mechanical waves require a physical medium to transmit energy.

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If the force provided by the engine of a toy car decreases, the car will slow down and eventually come to a stop.

1. The force provided by the engine of a toy car is responsible for propelling it forward at a constant speed. This force overcomes any friction or resistance acting on the car.

2. When the force provided by the engine decreases, there is a reduction in the overall force acting on the car. As a result, the force can no longer counteract the resistance and friction effectively.

3. The resistance and friction acting on the toy car, such as air resistance and the friction between the wheels and the surface, start to have a greater impact on the car's motion.

4. With a reduced force from the engine, the car begins to slow down gradually. The deceleration occurs because the opposing forces now have a greater influence on the car's motion.

5. As the force continues to decrease, the opposing forces eventually surpass the remaining force from the engine. Consequently, the toy car slows down even more and eventually comes to a complete stop.

6. If the force provided by the engine becomes extremely low or nonexistent, the opposing forces will completely overpower the forward force, causing the toy car to stop moving altogether.

In summary, when the force provided by the engine of a toy car decreases, the car's speed decreases, and it eventually comes to a stop due to the increased influence of opposing forces.

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Diving seabirds need to account for the bending of light because water has a higher refractive index than air.

When light passes through the air-water interface, it slows down and changes direction, causing the image of an object to appear distorted. This phenomenon is known as refraction. For diving seabirds, the refraction of light can make it challenging to accurately locate their prey when diving, which is essential for their survival.

To account for the bending of light, diving seabirds have evolved specialized eyes that allow them to adjust their focus and perceive images differently than humans. These adaptations enable the birds to see through the distorted image caused by refraction and accurately locate their prey while diving. One adaptation that diving seabirds have is a flattened cornea, which allows them to maintain a sharp focus while diving.

Overall, the ability of diving seabirds to account for the bending of light is essential for their survival, as it enables them to accurately locate and catch their prey while diving in the often murky and challenging conditions of the ocean.

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